Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
0 votes

Find $\sin \large\frac{x}{2},$$\cos \large\frac{x}{2}$ and $\tan \large\frac{x}{2}$ in each of the following $\tan x=-\large\frac{4}{3}$,$x$ in quadrant II

$\begin{array}{1 1} \sin \frac{x}{2} =\frac{2}{\sqrt 5} ; \cos \frac{x}{2} =\frac{-1}{\sqrt 5} \\ \sin \frac{x}{2} =\frac{-2}{\sqrt 5} ; \cos \frac{x}{2} =\frac{1}{\sqrt 5} \\ \sin \frac{x}{2} =\frac{-2}{\sqrt 5} ; \cos \frac{x}{2} =\frac{-1}{\sqrt 5} \\ \sin \frac{x}{2} =\frac{2}{\sqrt 5} ; \cos \frac{x}{2} =\frac{1}{\sqrt 5} \end{array}$

Can you answer this question?

1 Answer

0 votes
  • $\cos x=\large\frac{-1}{\sqrt{1\tan^2x}}$
Since $x$ lies in the second quadrant
$\therefore \cos x$ is negative.
$\therefore \cos x=\large\frac{-1}{\sqrt{1+\tan^2x}}=-\frac{1}{\sqrt{1+\Large\frac{16}{9}}}=-\frac{3}{5}$
Now ,$x$ lies in II quadrant
$\Rightarrow \large\frac{\pi}{2}$$ < x < \pi\Rightarrow \large\frac{\pi}{4} $$< \large\frac{x}{2}$$ < \large\frac{\pi}{2}$
$\Rightarrow \large\frac{x}{2}$ lies in first quadrant.
$\Rightarrow \sin \large\frac{x}{2}$$,\cos \large\frac{x}{2}$ and $\tan \large\frac{x}{2}$ are positive
$\therefore\cos\large\frac{x}{2}=\sqrt{\large\frac{1+\cos x}{2}}=\sqrt{\large\frac{1-\Large\frac{3}{5}}{2}}=\sqrt{\large\frac{5-3}{2\times 5}}=\frac{1}{\sqrt 5}$
$\therefore\sin\large\frac{x}{2}=\sqrt{\large\frac{1-\cos x}{2}}=\sqrt{\large\frac{1+\Large\frac{3}{5}}{2}}=\sqrt{\large\frac{5+3}{2\times 5}}=\frac{2}{\sqrt 5}$
$\therefore\tan\large\frac{x}{2}=\sqrt{\large\frac{1-\cos x}{1+\cos x}}=\sqrt{\large\frac{1+\Large\frac{3}{5}}{1-\Large\frac{3}{5}}}=\sqrt{\large\frac{5+3}{5-3}}=$$\sqrt{4}=2$
answered Apr 18, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App